LeetCode *** 129. Sum Root to Leaf Numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/treeshy/article/details/51078393

本文介绍了一种解决二叉树路径之和问题的方法，通过广度优先搜索遍历二叉树并累加从根节点到叶子节点的所有路径数值。

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题目：

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

分析：

这就是一道对二叉树进行遍历的题目，总体来说还是比较简单。我用的是广度优先遍历。。（深度优先宝宝我不是咋会。。。。）维护两个数组（一个是二叉树左右子树的数组，另外一个是相应位置的结果），大概就是这样。

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumNumbers(TreeNode* root) {
        
        if(root==NULL)return 0;
        
        int res=0;
        queue<int> tmpRes;
        queue<TreeNode *> tmpLeaf;
        
        tmpRes.push(res);
        tmpLeaf.push(root);
        
        while(tmpLeaf.size()!=0){
            TreeNode* tmpRoot=tmpLeaf.front();
            tmpLeaf.pop();
            int tmp=tmpRes.front();
            tmpRes.pop();
            
            tmp=tmp*10+tmpRoot->val;
            
            if(tmpRoot->left==NULL&&tmpRoot->right==NULL)res+=tmp;
            if(tmpRoot->left){
                tmpLeaf.push(tmpRoot->left);
                tmpRes.push(tmp);
            }
            if(tmpRoot->right){
                tmpLeaf.push(tmpRoot->right);
                tmpRes.push(tmp);
            }
        }
        
        return res;
        
    }
};